Timeline for Aren't qubits just ternary?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 28, 2021 at 20:10 | history | edited | Adam Zalcman | CC BY-SA 4.0 |
corrected range of theta mistakenly changed in previous edit (note that making the first amplitude positive (if possible) fixes the global phase)
|
| Feb 15, 2021 at 22:59 | comment | added | Quantum Guy 123 | OP might be wondering what the notation you are using means. To answer this question in layman terms: when a qubit is said to be "1 and 0 at the same time" what it means is that, it is in a probabilistic state. With some probability P, it is a 1, and with probability 1-P, it is 0. Additionally, qubits are also in a specific 'phase' which is analogous to phases in sin and cosin functions. | |
| Feb 13, 2021 at 21:37 | history | edited | user1271772 No more free time | CC BY-SA 4.0 |
added 75 characters in body
|
| Jan 27, 2021 at 17:58 | comment | added | MrArsGravis | More generally, in an $n$-qudit Hilbert space (= a Hilbert space with $n$ basis states), one can attach a coefficient $c_k \in \mathbb{C}$ to each of the basis states, leading to "as many different states" as $\mathbb{C}^n \cong \mathbb{R}^{2n}$ has, but one degree of freedom (DOF) is removed due to normalization of the state and one is removed since QM is invariant under global phase changes, i.e., one has $2n-2$ DOF, (or "infinity to the $2n-2$ different states", figuratively). | |
| Jan 26, 2021 at 20:12 | vote | accept | jort57 | ||
| Jan 26, 2021 at 20:10 | history | answered | Adam Zalcman | CC BY-SA 4.0 |